Introduction to Electrodynamics 4e

Published by Pearson Education
ISBN 10: 9332550441
ISBN 13: 978-9-33255-044-5

Chapter 1 - Section 2.7 - Differential Calculus - Problem - Page 24: 27

Answer

${\vec{\triangledown}}\cdot ({\vec{\triangledown}\times \vec{v_a}}) = 0$

Work Step by Step

Consider a vector function $\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$ To show $\vec{\triangledown}\cdot (\vec{\triangledown} \times \vec{F}) = 0$ Consider, $\vec{\triangledown} \times \vec{F} = \begin{vmatrix} \hat{i} && \hat{j} && \hat{k} \\ {\partial \over \partial x} && {\partial \over \partial y} && {\partial \over \partial z}\\ F_x && F_y && F_z \end{vmatrix}$ $\vec{\triangledown} \times \vec{F} = \hat{i}\big( {\partial \over \partial y} F_z - {\partial \over \partial z} F_y\big) + \hat{j}\big( {\partial \over \partial z} F_x - {\partial \over \partial x} F_z\big)+ \hat{k}\big( {\partial \over \partial x} F_y - {\partial \over \partial y} F_x\big) $ Now, $\vec{\triangledown}\cdot (\vec{\triangledown} \times \vec{F}) = {\partial \over \partial x}\big( {\partial \over \partial y} F_z - {\partial \over \partial z} F_y\big) + {\partial \over \partial y}\big( {\partial \over \partial z} F_x - {\partial \over \partial x} F_z\big) + {\partial \over \partial z}\big( {\partial \over \partial x} F_y - {\partial \over \partial y} F_x\big) = 0$ For $\vec{v_a} = x^2 \hat{i} + 3xz^2 \hat{j} − 2xz\hat{k}$ $\vec{\triangledown} \times \vec{v_a} = -6xz \hat{i} + 2z\hat{j} + 3z^2\hat{k}$ ${\vec{\triangledown}}\cdot ({\vec{\triangledown}\times \vec{v_a}}) = -6z + 0 + 6z = 0$ Hence. Divergence of curl is always zero.
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